Question

1) cosα=1/2 0(gradus)<α<90(gradus) sinα=? 2)sinα=-1/2 π<α<3π/2 cosα=? Даю 15б срочно помогите пожалуйста!!!!!

Answer 1

Воспользуемся тригонометрическим тождеством: $cos^2\alpha+sin^2\alpha$
1) $sin\alpha=\sqrt{1-cos^2\alpha}$
$\\sin\alpha=\sqrt{1-\frac{1}{4}}=\sqrt{\frac{3}{4}}=\frac{\sqrt{3}}{2}$
минус не стоит, т.к. угол находится в 1 четверти, а синус в нем положительный

2) $cos\alpha=-\sqrt{1-sin^2\alpha}=-\sqrt{1-\frac{1}{4}}=-\frac{\sqrt{3}}{2}$
- стоит перед корнем, т.к. угол находится в 3 четверти, а косинус в нем отрицательный

Answer 2

1. Given that cosα = 1/2 and 0° < α < 90°, we can find sinα using the identity sin²α + cos²α = 1.

We know that cosα = 1/2, and since α is in the first quadrant (0° < α < 90°), we can determine that sinα is positive.

Using the identity sin²α + cos²α = 1, we can substitute the given value of cosα and solve for sinα:

sin²α + (1/2)² = 1 sin²α + 1/4 = 1 sin²α = 1 - 1/4 sin²α = 3/4

Taking the square root of both sides (since sinα is positive):

sinα = √(3/4) sinα = √3/2

Therefore, sinα = √3/2.

2. Given that sinα = -1/2 and π < α < 3π/2, we can find cosα using the identity sin²α + cos²α = 1.

We know that sinα = -1/2, and since α is in the third quadrant (π < α < 3π/2), we can determine that cosα is negative.

Using the identity sin²α + cos²α = 1, we can substitute the given value of sinα and solve for cosα:

(-1/2)² + cos²α = 1 1/4 + cos²α = 1 cos²α = 1 - 1/4 cos²α = 3/4

Taking the square root of both sides (since cosα is negative):

cosα = -√(3/4) cosα = -√3/2

Therefore, cosα = -√3/2.